codeforces-73C. LionAge II

传送门

一个小写字母组成的字符串和允许更改次数k，n组序偶的价值，求能获得的最大价值

定义dp[k][l][i]位更改第k次时长度为l的串末尾为i时的最大收益。分原本字符串末尾就是i和不是两种去递推。

注意初始化问题。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int maxn = 1e2 + 10;
const int maxk = 110;
char str[maxn];
int val[30][30];
int N, K;
int dp[maxk][maxn][30];

int main() {
    scanf("%s", str);
    int len = strlen(str);
    scanf("%d", &K);
    scanf("%d", &N);
    char c1, c2;
    int v;
    while (N--) {
        getchar();
        scanf("%c %c %d", &c1, &c2, &v);
        val[c1 - 'a'][c2 - 'a'] = v;
    }
    for (int k = 0; k <= K; k++) {
        for (int l = 0; l <= len; l++) {
            for (int i = 0; i < 26; i++) {
                dp[k][l][i] = -INF;
            }
        }
    }
    for (int i = 0; i < 26; i++) dp[1][1][i] = 0;
    dp[0][1][str[0] - 'a'] = 0;
    for (int i = 2; i <= len; i++) {
        int t1 = str[i - 1] - 'a', t2 = str[i - 2] - 'a';
        dp[0][i][t1] = dp[0][i - 1][t2] + val[t2][t1];
    }
    for (int l = 2; l <= len; l++) {
        for (int k = 1; k <= K; k++) {
            for (int i = 0; i < 26; i++) {
                for (int j = 0; j < 26; j++) {
                    if (str[l - 1] - 'a' == j) dp[k][l][j] = max(val[i][j] + dp[k][l - 1][i], dp[k][l][j]);
                    if (dp[k - 1][l - 1][i] == -INF) continue;
                    dp[k][l][j] = max(dp[k][l][j], dp[k - 1][l - 1][i] + val[i][j]);
                }
            }
        }
    }
    int ans = -INF;
    for (int i = 0; i <= K; i++) {
        for (int j = 0; j < 26; j++) {
            ans = max(ans, dp[i][len][j]);
        }

    }
    printf("%d
", ans);
    return 0;
}